Edexcel A-Level Chemistry AS Paper 2, June 2024: Question 8

9 marks · Medium difficulty · Practical Techniques and Data Analysis

Write the combustion equation for methanol, calculate the theoretical final temperature of water in a calorimetry experiment, and evaluate the effect of balance precision on accuracy.

Practise this question

Question

Question 8 begins with a diagram showing a calorimetry setup: a spirit burner containing methanol under a beaker of water held by a clamp, with a thermometer immersed in the water. A table of student results shows mass of burner before combustion (152.2 g), after combustion (150.2 g), mass of water (200.0 g), and water temperatures before (20.5 °C) and after heating (52.5 °C). Given data includes enthalpy of combustion of methanol (-726 kJ mol⁻¹), specific heat capacity of water (4.18 J g⁻¹ °C⁻¹), and molar mass of methanol (32.0 g mol⁻¹). Part (a)(i) asks for the equation representing the enthalpy change of combustion of methanol with state symbols (2 marks). Part (a)(ii) asks to calculate the expected final temperature of the water assuming no experimental errors (4 marks). Part (a)(iii) asks to calculate the percentage error in the experimental temperature rise compared to the theoretical temperature rise (1 mark). Part (b) asks to explain whether using a 2 decimal place balance instead of a 1 decimal place balance would have given a final answer with greater accuracy (2 marks).
Question text

8 A student carried out an experiment to determine the enthalpy change of

combustion of methanol.

Diagram

thermometer

beaker

water

spirit burner

methanol

Student’s results

Measurement Value

Mass of spirit burner and methanol before combustion 152.2g

Mass of spirit burner and methanol after combustion 150.2g

Mass of water in the beaker 200.0g

Temperature of water before heating 20.5°C

Temperature of water after heating 52.5°C

Data

The enthalpy change of combustion of methanol, Δ H = –726 kJ mol–1

c

The specific heat capacity of water, c = 4.18 J g–1 °C–1

Molar mass of methanol = 32.0 g mol–1

(a) (i) Write the equation to represent the enthalpy change of combustion of

methanol. Include state symbols.

(2)

(ii) Calculate the expected final temperature of the water in the student’s

experiment, assuming no experimental errors.

(4)

*P76894A02632*

(iii) Calculate the percentage error in the experimental temperature rise

compared to the theoretical temperature rise from your calculation.

(1)

(b) Instead of waiting in a queue for a balance which recorded the mass of the

spirit burner to 2 decimal places, the student used a balance which recorded the

mass to 1 decimal place.

Explain whether or not the student would have been better to wait for the

balance with greater precision to give a final answer with greater accuracy.

(2)

… *P76894A02732*

(Total for Question 8 = 9 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 8: (a)(i) requires correct substances CH3OH(l) + 1.5O2(g) -> CO2(g) + 2H2O(l) (1 mark) and balancing with state symbols without multiples (1 mark). (a)(ii) awards 1 mark for moles of methanol burnt = (152.2 - 150.2) / 32.0 = 0.0625 mol; 1 mark for heat energy = 726 * 0.0625 = 45.375 kJ (45375 J); 1 mark for temperature change = 45.375 / (200 * 0.00418) = 54.3 °C; 1 mark for final temperature = 54.3 + 20.5 = 74.8 °C. (a)(iii) awards 1 mark for percentage error = ((52.5 - 20.5) - 54.3) / 54.3 * 100 = (-)41%. Part (b) awards 1 mark for noting the balance uncertainty is very small compared to experimental percentage error, and 1 mark for stating that heat loss to surroundings causes much bigger uncertainty/error, so higher balance precision does not improve accuracy.

Question

Answer Additional Guidance Mark

Number

8(a)(i) (2)

• substances (1) CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)

• state symbols and balancing (1) Do not award multiples

Question

Answer Additional Guidance Mark

Number

8(a)(ii) Example of calculation (4)

• calculation of moles of methanol burnt (1) (152.2 ‒ 150.2) ÷ 32.0 = 0.0625 (mol)

• calculation of expected energy transferred (1) 726 × 0.0625 = 45.375 (kJ)

by multiplying by number of moles

and Award

scaling to match units 726000 × (ans 1) = 45375 (J)

• calculation of temperature change by 45.375 ÷ (200 × 0.00418) = 54.276 / 54.3 (ºC)

dividing by 836 (1) / 45.375 ÷ 0.836 = 5 4.276 / 54.3 (ºC)

Award

45375 ÷ (200 × 4.18) = 54.276 / 54.3 (ºC)

45375 ÷ 836 = 54.276 / 54.3 (ºC)

• calculation of expected final temperature by 54.3 + 20.5 = 74.8 (ºC)

adding 20.5 (1)

Allow TE throughout

Ignore SF except 1 SF

Question

Answer Additional Guidance Mark

Number

8(a)(iii) Example of calculation (1)

• calculation of percentage error experimental temperature rise – theoretical temperature rise × 100 =

theoretical temperature rise

((52.5 ‒ 20.5) ‒ 54.3) ÷ 54.3 × 100 = (‒) 41.068 / (‒) 41%

32 – 54.3 = 22.3

22.3 ÷ 54.3 × 100 = (‒) 41.068 / (‒) 41%

Allow (54.3 ‒ (52.5 ‒ 20.5)) ÷ 54.3 × 100 = 41.068 / 41%

Allow TE on temperature rise from (a)(ii)

Ignore SF except 1SF

Question

Answer Additional Guidance Mark

Number

An answer that makes reference to the following points: (2)

• (there is no need to use a balance with greater precision because)

the uncertainty of the 1 dp balance is very small (compared to the

percentage error)

(1)

• (in thermochemistry experiments the random error associated with)

the heat lost to the surrounding causes a much bigger uncertainty in

the final value (than the balance)

or

the increased precision will not lead to greater accuracy /

the accuracy of the experiment is poor so a high degree of precision Allow the perentage error in the

will not give a better result temperature change is so high it would not

(1) lead to greater accuracy

(Total for Question 8 = 9 marks)

TOTAL FOR PAPER = 80 MARKS

How to answer it

Enthalpy Change of Combustion of Methanol

📌 WHAT THIS QUESTION TESTS

This question assesses practical thermochemistry and experimental analysis:

  • Writing standard thermochemical equations with correct definitions and state symbols.
  • Performing reverse calorimetry calculations ( q = mcΔT ) to determine theoretical temperature rise and final temperature.
  • Evaluating experimental data by calculating percentage error.
  • Critically analysing experimental design, distinguishing between apparatus precision and systematic experimental errors (such as heat loss).
PART (a)(i) — 2 MARKS

Standard Enthalpy of Combustion Equation

Writing the balanced chemical equation for the standard enthalpy change of combustion of methanol

✅ Correct Answer

CH₃OH(l) + 1.5O₂(g) → CO₂(g) + 2H₂O(l)

Also accepted for oxygen: 1½O₂(g) or ³⁄₂O₂(g)

Mark 1: Correct species ( CH₃OH + O₂ → CO₂ + H₂O )
Mark 2: Fully balanced for 1 mole of methanol and correct state symbols.

💡 Key Knowledge

  • Standard enthalpy of combustion (ΔcH⦵): The enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions (298 K, 100 kPa).
  • Because it is defined per 1 mole of fuel, the coefficient of CH₃OH must be 1.
  • Standard states at 298 K: methanol is a liquid (l) , water is a liquid (l) , oxygen and carbon dioxide are gases (g) .

❌ Common Errors

  • Doubling the equation: 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O gets 0/2 for balancing because ΔcH must refer to 1 mole of fuel.
  • Wrong state of water: Writing H₂O(g) loses the second mark. Standard state for water at 298 K is liquid.
  • Missing methanol's state: Methanol is liquid (l) , not aqueous (aq) .

🧠 Exam Technique

Whenever you see "equation to represent the enthalpy change of combustion", write the fuel first with a coefficient of 1, balance the carbons into CO₂, hydrogens into H₂O, and finally balance the oxygen atoms—remembering the oxygen atom already inside the alcohol molecule!

PART (a)(ii) — 4 MARKS

Calculating Expected Final Temperature

Working backwards from ΔcH to find the theoretical final water temperature

📐 Step-by-Step Calculation

1 Find mass and moles of methanol burned:
Mass burnt = 152.2 g - 150.2 g = 2.0 g
Moles of CH₃OH = 2.0 / 32.0 = 0.0625 mol [1 Mark]

2 Calculate theoretical energy released (q):
q = moles × |ΔcH| = 0.0625 mol × 726 kJ mol⁻¹ = 45.375 kJ
Convert to Joules (J): 45.375 × 1000 = 45375 J [1 Mark]

3 Calculate theoretical temperature rise (ΔT):
Using q = m × c × ΔT , rearrange to ΔT = q / (m × c)
Note: m is the mass of water heated = 200.0 g
ΔT = 45375 / (200.0 × 4.18) = 45375 / 836 = 54.276 °C (or 54.3 °C) [1 Mark]

4 Calculate expected final temperature:
Final Temperature = Initial Temperature + ΔT
Final Temperature = 20.5 + 54.276 = 74.8 °C (accepts 74.78 °C) [1 Mark]

❌ Common Errors & Traps

  • Stopping at ΔT: Many students calculated 54.3 °C and stopped, forgetting the question asked for the final temperature, losing the 4th mark.
  • Unit clash in q = mcΔT: Forgetting to multiply 45.375 kJ by 1000 before dividing by c (4.18 J g⁻¹ °C⁻¹).
  • Using wrong mass in q = mcΔT: Using 2.0 g (mass of fuel) instead of 200.0 g (mass of water being heated).

🧠 Exam Technique

Transfer of Error (TE) is applied throughout this multi-step question. Even if you miscalculated moles in Step 1, carrying that number forward correctly through the remaining steps still secures up to 3 out of 4 marks.

PART (a)(iii) — 1 MARK

Percentage Error in Temperature Rise

Evaluating the experimental temperature rise against theoretical predictions

📐 Calculation

1. Experimental temperature rise:
ΔTexp = 52.5 °C - 20.5 °C = 32.0 °C

2. Theoretical temperature rise:
ΔTtheor = 54.276 °C (from part (a)(ii))

3. Percentage error:
|ΔTexp - ΔTtheor| / ΔTtheor × 100
= |32.0 - 54.276| / 54.276 × 100
= 22.276 / 54.276 × 100 = 41.0% (or 41%)

❌ Common Errors

  • Dividing by experimental instead of theoretical: Percentage error relative to theory must always have the theoretical (true) value in the denominator.
  • Using final temperatures instead of rises: The question specifies percentage error in temperature rise, not the final temperature.
Mark 1: Percentage error calculation = 41% or 41.1% (or -41%, allow TE from (a)(ii)).
PART (b) — 2 MARKS

Apparatus Precision vs. Experimental Accuracy

Evaluating whether a 2 decimal place balance would meaningfully improve the result

✅ Model Answer

The student would not have been significantly better off waiting for the 2 d.p. balance:

  • The apparatus uncertainty/percentage error of the 1 d.p. balance (±0.05 g on 2.0 g = ~5%) is very small compared to the huge overall experimental percentage error (~41%). [1 Mark]
  • The primary limitation on accuracy is heat loss to the surroundings (or incomplete combustion/evaporation of fuel), which creates a much larger systematic error than balance precision. [1 Mark]

💡 Precision vs. Accuracy

  • Precision: How close repeated measurements are to each other, influenced by the resolution of apparatus (e.g., 1 d.p. vs 2 d.p. balance).
  • Accuracy: How close a measurement is to the true theoretical value.
  • Improving instrument precision cannot fix large systematic errors like heat loss to the air and beaker.

🧠 Examiner Insight

Top students recognized that an experimental error of >40% cannot be improved by changing a balance that only has an uncertainty of a few percent. Always compare the magnitude of apparatus uncertainty with experimental discrepancy.

❌ What Lost Marks

Saying simply "Yes, because 2 d.p. is more accurate" without referencing the context of heat loss or comparing the error magnitudes scored 0/2.

Topics

Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.